#include <iostream>
#include <vector>

using namespace std;

/**
 * 状态：f[i][j] : s的前i个字符是否匹配 p的前j个字符
 * 执行用时：8 ms, 在所有 C++ 提交中击败了76.63%的用户
 * 内存消耗：7.2 MB, 在所有 C++ 提交中击败了49.69%的用户
 */
class Solution {
public:
    bool isMatch(string s, string p) {

        int n = s.size(), m = p.size();
        s  = ' ' + s, p = ' ' + p;
        vector<vector<bool>> f(n + 1, vector<bool>(m + 1));
        f[0][0] = true;
        for (int i = 0; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                if (j + 1 <= m && p[j + 1] == '*') continue;
                if (i && p[j] != '*') {
                    f[i][j] = f[i - 1][j - 1] && (s[i] == p[j] || p[j] == '.');
                } else if (p[j] == '*') {
                    f[i][j] = f[i][j - 2] || (i && f[i - 1][j] && (s[i] == p[j - 1] || p[j - 1] == '.'));
                }
            }
        }

        return f[n][m];
    }
};

int main() {

    cout << Solution().isMatch("aa", "a") << endl;  // false
    cout << Solution().isMatch("aa", "a*") << endl;  // true
    cout << Solution().isMatch("ab", ".*") << endl;  // true
    cout << Solution().isMatch("aab", "c*a*b") << endl;  // true

    return 0;
}
